If ($$sin^{2}36^\circ+tan^{2}60^\circ+sec^{2}30^\circ+sin^{2}54^\circ$$)is equal to:
$$sin^{2}36^\circ+tan^{2}60^\circ+sec^{2}30^\circ+sin^{2}54^\circ=sin^{2}36^\circ+tan^{2}60^\circ+sec^{2}30^\circ+sin^{2}(90-36)^\circ$$
$$=sin^{2}36^\circ+cos^{2}36^\circ+tan^{2}60^\circ+sec^{2}30^\circ$$
$$=1+\left(\sqrt{3}\right)^2+\left(\frac{2}{\sqrt{3}}\right)^2$$
$$=1+3+\frac{4}{3}$$
$$=\frac{16}{3}$$
Hence, the correct answer is Option C
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